Question

show that √2 is a irrational number​

Answer 1

AnswEr :

$\normalsize\sf\ Let \: \red{\sqrt{2}} \; be \: a \: rational \: number$

$\normalsize\sf\ As; \: we \: know \: Rational \: number \: is \: in \: form \: of \: \green{\frac{p}{q}}$

$\normalsize\dashrightarrow\sf\red{\sqrt{2} } = \green{\frac{a}{b}}$

$\normalsize\quad\sf\ [\because\ a \: and \: b \: are \: co-prime \: and \: b \: \neq\ \: 0]$

$\underline{\bigstar\:\textsf{Squaring \: both \: sides( L.H.S \& R.H.S):}}$

$\normalsize\ : \implies\sf\ (\sqrt{2})^2 = \left[\frac{p}{q} \right]^2$

$\normalsize\ : \implies\sf\ 2 = \frac{a^2}{b^2}$

$\normalsize\ : \implies\sf\ a^2 = 2b^{2} \: \: ---(eq.1)$

$\normalsize\ : \implies\sf\frac{a^2}{2} = b^{2}$

$\normalsize\sf\ \therefore\ a^{2} \: is \: divisible \: by \: 2 \: and \: a \: is \: also \: divisible \: by \: 5$

$\rule{170}1$

$\normalsize\sf\ Now;$

$\normalsize\: \implies\sf\ a = 2c$

$\normalsize\qquad\sf\ [\because\ Where \: k \: is \: an \: integer]$

$\underline{\bigstar\:\textsf{Squaring \: both \: sides( L.H.S \& R.H.S):}}$

$\normalsize\ : \implies\sf\ a^2 = (2c)^{2}$

$\scriptsize\tt{\quad\dag\ put \: the \: value \: of \: b^2 \: from \: eq.1}$

$\normalsize\ : \implies\sf\ \cancel{2}b^2 = \cancel{4}c^{2}$

$\normalsize\ : \implies\sf\ b^2 = 2c^{2}$

$\normalsize\ : \implies\sf\frac{b^2}{2} = c^{2}$

$\normalsize\sf\ \therefore\ b^{2} \: is \: divisible \: by \: 2 \: and \: b \: is \: also \: divisible \: by \: 5$

$\normalsize\sf\ This \: shows \: that \: \green{a} \: and \: \green{b} \: have \: common \: factor \: as \: 5 \\ \normalsize\sf\ but \: if \: they \: are \: \blue{rational} \: they \: are \: \pink{co-prime}.$

$\normalsize\sf\ This \: contradicts \: the \: fact \: that \: \red{\sqrt{2}} \: is \: \blue{rational}$

$\normalsize\sf\ Hence, \: Our \: assumption \: is \: wrong \: \\ \normalsize\sf\ \red{\sqrt{2}} \: is \: \pink{irrational}$

$\large\maltese \: \: {\boxed{\sf \orange{Hence \: Prove \: !!}}}$

Answer 2

Let us assume that √2 is a rational number.

So it will be expressed in the form of $\red{ \frac{p}{q} }$

Where q is not equal to 0.

Here p and q are co primes whose HCF is 1.

${ \red{\sqrt{2} = \frac{p}{q} }}$

Squaring both sides, we get

$2 = \frac{ {p}^{2} }{ \ {q}^{2} } \\ \implies2 {q }^{2} = {p}^{2} ........(1) \\ \implies \: 2 \: divides \: {p}^{2} \\ \implies \: 2 \: divides \: p$

Here 2 divides p.

Now let p= 2c

Squaring both sides,

${p}^{2} = 4 {c}^{2} .......(2)$

Substituting (1) into (2)

$\implies2 {q}^{2} = 4c^{2} \\ \implies \: {q}^{2} = 2 {c}^{2} \\ \implies \: 2 {c}^{2} = {q}^{2}$

Here 2 divides q² as well as q.

Here p and q are Divisible by 2.

So, our assumption is wrong.

Hence √2 is irrational number.